Metamath Proof Explorer


Theorem mhmlin

Description: A monoid homomorphism commutes with composition. (Contributed by Mario Carneiro, 7-Mar-2015)

Ref Expression
Hypotheses mhmlin.b B=BaseS
mhmlin.p +˙=+S
mhmlin.q ˙=+T
Assertion mhmlin FSMndHomTXBYBFX+˙Y=FX˙FY

Proof

Step Hyp Ref Expression
1 mhmlin.b B=BaseS
2 mhmlin.p +˙=+S
3 mhmlin.q ˙=+T
4 eqid BaseT=BaseT
5 eqid 0S=0S
6 eqid 0T=0T
7 1 4 2 3 5 6 ismhm FSMndHomTSMndTMndF:BBaseTxByBFx+˙y=Fx˙FyF0S=0T
8 7 simprbi FSMndHomTF:BBaseTxByBFx+˙y=Fx˙FyF0S=0T
9 8 simp2d FSMndHomTxByBFx+˙y=Fx˙Fy
10 fvoveq1 x=XFx+˙y=FX+˙y
11 fveq2 x=XFx=FX
12 11 oveq1d x=XFx˙Fy=FX˙Fy
13 10 12 eqeq12d x=XFx+˙y=Fx˙FyFX+˙y=FX˙Fy
14 oveq2 y=YX+˙y=X+˙Y
15 14 fveq2d y=YFX+˙y=FX+˙Y
16 fveq2 y=YFy=FY
17 16 oveq2d y=YFX˙Fy=FX˙FY
18 15 17 eqeq12d y=YFX+˙y=FX˙FyFX+˙Y=FX˙FY
19 13 18 rspc2v XBYBxByBFx+˙y=Fx˙FyFX+˙Y=FX˙FY
20 9 19 syl5com FSMndHomTXBYBFX+˙Y=FX˙FY
21 20 3impib FSMndHomTXBYBFX+˙Y=FX˙FY